Question 1)Read the data from one file and remove all the
duplicated words and generate an output file containg only
using words.Program should also print number of occurance
of each words on standard output.(Program must be use STL
and iostream object).
Question 2)Write a program to convert the lower case
contents of file to upper case using STL.
Question 3)What is the output of this problem.
int i=12;
int &r =i;
r+r/4;
int +p =&r;
int *p = &r;
P+=r;
return 1;
Answer :a)12 b)17 c) 30 d)24 E)15
Question 4)
#include
#include
#include
void main()
{
char srcstr[30],desstr[30];
int i,len;
clrscr();
cout<<"\nenter the string\n";
cin>>srcstr;
len=strlen(srcstr);
for(i=0;srcstr[i]!='\0';i++)
{
desstr[--len]=srcstr[i];
}
desstr[i]='\0';
cout<<"\nreversed string is\n";
for(i=0;desstr[i]!='\0';i++)
{
cout< }
getch();
}
Answer :
A) string output b)tuptuo gnirts c) string d)output
Question 5)
Class Test
{
static const Tk=LEN;
int q;
public:
foo(intx =LEN):q(x){};
TGet Q() const { return q:}
void call test
Foo<long :9> foo(10);
std::const<< foo.Getq();
Answer: 7,8,9,10,11
question 6)
class A
{
int a;
char b;
};
class B:public A
{
char b;
int a;
};
What is the size?
Answer
a)5 bytes for class A and 5 bytes class B
b)4 bytes for class A and 8 bytes class B.
c)....
d)....
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Answer / kaustubh
Q2)Solution Code
#include <iostream>
#include <fstream>
#include <list>
using namespace std;
int main()
{
ifstream in("input.txt",ios::in);
ofstream out("output.txt",ios::out);
list<char> lst;
list<char>::iterator p;
char ch;
if(!in)
{
cout<<"Cannot open input.txt";
exit(0);
}
while(in)
{
in.get(ch);
lst.push_back(ch);
}
p=lst.begin();
while(p!=lst.end())
{
*p=toupper(*p);
p++;
}
p=lst.begin();
while(p!=lst.end())
{
out<<*p;
p++;
}
in.close();
out.close();
}
| Is This Answer Correct ? | 8 Yes | 0 No |
Answer / manoj kumar kar
Question 3)What is the output of this problem.
int i=12;
int &r =i;
r+r/4;
int *p = &r;
P+=r;
return i;
Answer :a)12 b)17 c) 30 d)24 E)15
// Answer will be 12.
r is a reference to i so both r and i will be 12 at first.
r+r/4 is meaningless as there is no assinment.
p is a pointer so increamenting p will not manipulate the
value stored at p(The value of r will not be changed if we
increase p if we want to increase the value of r then we
have to do (*p)+=r).
Qustion 4:
//It will output the reverse string.
Question 5:
Not clear with the question.
Question 6:
//Space is always reserved for base class variables in
derived class objects.So if sizeof(int)=4 and sizeof(char)=1
then answer will be
5 Bytes for Class A and 10 Bytes for Class B.
| Is This Answer Correct ? | 6 Yes | 0 No |
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